The setup is as follows: Let $X/\mathbb{C}$ be a curve, and let $t : X\rightarrow\mathbb{P}^1_\mathbb{C}$ be a meromorphic function on $X$ thought of as a covering map of degree $n$. Further, let $\text{Crit}(t)$ denote the critical points of the cover $t$ - ie, the points in $\mathbb{P}^1_\mathbb{C}$ which have fewer than $n$ pre-images under $t$.

Now, it's my understanding that the critical points of $t$ should be the images of the ramification points of $t$ under $t$, so I've been trying to understand why it should be the case that the sheaf of relative differentials of $X/\mathbb{P}^1_\mathbb{C}$ should be nonzero only on the ramification points (or at least only above the critical points).

To this end, I'm trying to understand the definition given in Hartshorne (III.8), namely:
$\Omega^1_{X/\mathbb{P}^1_\mathbb{C}} = \Delta^*(\mathcal{I}/\mathcal{I}^2)$, where $\Delta : X\rightarrow X\times_{\mathbb{P}^1_\mathbb{C}} X$ is the diagonal map, and $\mathcal{I}$ is the sheaf of ideals of the image $\Delta(X)$ in some open subset $W\subset X\times_{\mathbb{P}^1_\mathbb{C}} X$.

I kind of understand sheaves of ideals (they're essentially functions on the ambient space that vanish on the closed subscheme), but I'm still not very comfortable with the notion of $\Delta^*(\mathcal{I}/\mathcal{I}^2)$ (in this case defined to be $\Delta^{-1}\mathcal{I}/\mathcal{I}^2\otimes_{\Delta^{-1}\mathcal{O}_{X\times X}}\mathcal{O}_X$, where the fibred product is taken over $\mathbb{P}^1$).

Any comments on how I should think of $\Delta^*(...)$ and why the sheaf of relative differentials only have nonzero stalks at ramification points would be awesome!

Try looking at the local picture. Pick local coordinates on $X$ and $\mathbb P^1$, centered on a ramification point on $X$, such that the morphism is given by $z \mapsto z^k$. Write down the short exact sequence associated of tangent sheaves associated to the morphism, and note that the differential of $f$ is $k z^{k-1} dz$. For a fixed $z \not= 0$ this morphism $T_X \to f^*T_{\mathbb P^1}$ is surjective, because it is injective and both spaces are of dimension 1, so the sheaf of relative differentials is zero. For $z = 0$, the morphism is zero, whence the support of the relative sheaf.
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Gunnar Þór MagnússonApr 29 '12 at 7:58

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In my understanding, $\Delta^*(\mathcal{I}/\mathcal{I}^2)$ is technically convenient because it is obviously well-defined, but not a good way to reason about differentials in practice. First, you need to understand $\Omega_{X/\Bbbk}$, where $X$ is a smooth variety over a field $\Bbbk$. (This is essentially the sheaf of differential forms.) Once you understand this, $\Omega_{X/Y}$ is obtained (for a morphism $X \to Y$) by taking $\Omega_{X/\Bbbk}$ and modding out by pullbacks of differential forms on $Y$.
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Charles StaatsApr 29 '12 at 21:48

Qualification: $\Delta^*(\mathcal{I}/\mathcal{I}^2)$ is often not the best way to reason about differentials in practice, particularly when you are learning.
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Charles StaatsApr 29 '12 at 21:50

2 Answers
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Ravi Vakil has a good explaination for the definition $\Delta^*(I/I^2)$ in his notes. See his AG notes here or here (chapter 23). In particular, I guess thinking about this locally makes it a little clearer what's going on, in terms of derivations etc. Also, when $X$ is smooth, it is instructive to see that this really gives the cotangent bundle on $X$.

As for your question about ramification points: Let $f:X\to Y$ be a finite morphism of curves (I will assume that these are smooth in the following). It is useful to have in mind the exact sequence $$0\to f^*\Omega_{Y}\to \Omega_X \to \Omega_{X|Y}\to 0.$$(This is exact at the right in the smooth case, but not in general). Note that $\Omega_{X|Y}$ is a torsion sheaf since the two other sheaves are locally free of the same rank (they are line bundles on $X$). At a point $q\in Y$ and $p\in X$ in the preimage of $q$, let $dx$ denote a generator for $\Omega_{Y,q}$ as a $O_{Y,q}$-module. Now, $(\Omega_{X|Y})_P=0$ if and only if $f^*dx$ is a generator of $\Omega_{X,p}$, which happens if and only if $f$ pulls back a local parameter to a local parameter, that is $p$ is unramified. Moreover, the exact sequence above shows that the ramification index is exactly the length of the sheaf $\Omega_{X|Y}$. Finally, note that this sequence gives the Riemann hurwitz formula, relating the canonical divisors of $X$ and $Y$ and the ramification divisor of $f$.

Then, for any point $x$ in $X$ lying over $y$ in $Y$, the coefficient $v_x(\pi)$ of $\Omega_{\pi}$ is the valuation of the different of the extension of dvr's $\mathcal{O}_{y}\subset \mathcal{O}_x$. If you are working in characteristic zero, then $$v_y(\pi) = e_x-1,$$ where $e_x$ is the ramification index. So you see that $\Omega_{\pi}$ is supported on the ramification points.

Also, you have a short exact sequence (it's on page 2 of Chapter IV.2 in Hartshorne) which relates $\Omega_\pi$ with $\Omega_X$ and $\Omega_Y$. The above actually shows the important Riemann-Hurwitz formula: $$K_X = \pi^{\ast} K_Y + R.$$ Here $R$ is the ramification divisor. This equals $\Omega_{\pi}$ in this case.